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Date	May 2022	Marks available	1	Reference code	22M.2.hl.TZ1.1
Level	HL	Paper	2	Time zone	TZ1
Command term	Evaluate	Question number	1	Adapted from	N/A

Question

When heated in air, magnesium ribbon reacts with oxygen to form magnesium oxide.

The reaction in (a)(i) was carried out in a crucible with a lid and the following data was recorded:

Mass of crucible and lid = 47.372 ±0.001 g

Mass of crucible, lid and magnesium ribbon before heating = 53.726 ±0.001 g

Mass of crucible, lid and product after heating = 56.941 ±0.001 g

When magnesium is burnt in air, some of it reacts with nitrogen to form magnesium nitride according to the equation:

3 Mg (s) + N₂(g) → Mg₃N₂(s)

The presence of magnesium nitride can be demonstrated by adding water to the product. It is hydrolysed to form magnesium hydroxide and ammonia.

Most nitride ions are ¹⁴N^3–.

Write a balanced equation for the reaction that occurs.

[1]

a(i).

Identify a metal, in the same period as magnesium, that does not form a basic oxide.

[1]

a(ii).

Calculate the amount of magnesium, in mol, that was used.

[1]

b(i).

Determine the percentage uncertainty of the mass of product after heating.

[2]

b(ii).

Assume the reaction in (a)(i) is the only one occurring and it goes to completion, but some product has been lost from the crucible. Deduce the percentage yield of magnesium oxide in the crucible.

[2]

b(iii).

Evaluate whether this, rather than the loss of product, could explain the yield found in (b)(iii).

[1]

c(i).

Suggest an explanation, other than product being lost from the crucible or reacting with nitrogen, that could explain the yield found in (b)(iii).

[1]

c(ii).

Calculate coefficients that balance the equation for the following reaction.

[1]

d(i).

Ammonia is added to water that contains a few drops of an indicator. Identify an indicator that would change colour. Use sections 21 and 22 of the data booklet.

[1]

d(ii).

Determine the oxidation state of nitrogen in Mg₃N₂ and in NH₃.

[1]

d(iii).

Deduce, giving reasons, whether the reaction of magnesium nitride with water is an acid–base reaction, a redox reaction, neither or both.

[2]

d(iv).

State the number of subatomic particles in this ion.

[1]

e(i).

Some nitride ions are ¹⁵N^3–. State the term that describes the relationship between ¹⁴N^3– and ¹⁵N^3–.

[1]

e(ii).

The nitride ion and the magnesium ion are isoelectronic (they have the same electron configuration). Determine, giving a reason, which has the greater ionic radius.

[1]

e(iii).

Suggest, giving a reason, whether magnesium or nitrogen would have the greater sixth ionization energy.

[1]

e(iv).

Suggest two reasons why atoms are no longer regarded as the indivisible units of matter.

[2]

State the types of bonding in magnesium, oxygen and magnesium oxide, and how the valence electrons produce these types of bonding.

[4]

Markscheme

2 Mg(s) + O₂(g) → 2 MgO(s) ✔

Do not accept equilibrium arrows. Ignore state symbols

a(i).

aluminium/Al ✔

a(ii).

$⟨ ⟨ \frac{53.726 g - 47.372 g}{244.31 g {mol}^{- 1}} = \frac{6.354 g}{24.31 g {mol}^{- 1}} ⟩ ⟩ = 0.2614 « mol » ✔$

b(i).

mass of product $« = 56.941 g - 47.372 g » = 9.569 « g »$ ✔

$⟨⟨100 × \frac{2 \times 0.001 g}{9.569 g} =0.0209⟩⟩ = 0.02 «%»$ ✔

Award [2] for correct final answer

Accept 0.021%

b(ii).

$⟨ ⟨ 0.2614 mol \times (24.31 g {mol}^{- 1} + 16.00 g {mol}^{- 1}) = 0.2614 mol \times 40.31 g {mol}^{- 1} ⟩ ⟩ = 10.536 « g »$ ✔

$⟨ ⟨ 100 \times \frac{9.569 g}{10.536 g} = 90.822 ⟩ ⟩ = 91 « % »$ ✔

Award «0.2614 mol x 40.31 g mol^–1»

Accept alternative methods to arrive at the correct answer.

Accept final answers in the range 90.5-91.5%

[2] for correct final answer.

b(iii).

yes
AND
«each Mg combines with $\frac{2}{3}$ N, so» mass increase would be 14x $\frac{2}{3}$ which is less than expected increase of 16x
OR
3 mol Mg would form 101g of Mg₃N₂ but would form 3 x MgO = 121 g of MgO
OR
0.2614 mol forms 10.536 g of MgO, but would form 8.796 g of Mg₃N₂ ✔

Accept Yes AND “the mass of N/N₂ that combines with each g/mole of Mg is lower than that of O/O₂”

Accept YES AND “molar mass of nitrogen less than of oxygen”.

c(i).

incomplete reaction
OR
Mg was partially oxidised already
OR
impurity present that evaporated/did not react ✔

Accept “crucible weighed before fully cooled”.

Accept answers relating to a higher atomic mass impurity consuming less O/O₂.

Accept “non-stoichiometric compounds formed”.

Do not accept "human error", "wrongly calibrated balance" or other non-chemical reasons.

If answer to (b)(iii) is >100%, accept appropriate reasons, such as product absorbed moisture before being weighed.

c(ii).

«1» Mg₃N₂ (s) + 6 H₂O (l) → 3 Mg(OH)₂ (s) + 2 NH₃ (aq) ✔

d(i).

phenol red ✔

Accept bromothymol blue or phenolphthalein.

d(ii).

Mg₃N₂: -3
AND
NH₃: -3 ✔

Do not accept 3 or 3-

d(iii).

Acid–base:
yes AND N^3- accepts H⁺/donates electron pair«s»
OR
yes AND H₂O loses H⁺ «to form OH^-»/accepts electron pair«s» ✔

Redox:
no AND no oxidation states change ✔

Accept “yes AND proton transfer takes place”

Accept reference to the oxidation state of specific elements not changing.

Accept “not redox as no electrons gained/lost”.

Award [1 max] for Acid–base: yes AND Redox: no without correct reasons, if no other mark has been awarded

d(iv).

Protons: 7 AND Neutrons: 7 AND Electrons: 10 ✔

e(i).

isotope«s» ✔

e(ii).

nitride AND smaller nuclear charge/number of protons/atomic number ✔

e(iii).

nitrogen AND electron lost from first «energy» level/s sub-level/s-orbital AND magnesium from p sub-level/p-orbital/second «energy» level
OR
nitrogen AND electron lost from lower level «than magnesium» ✔

Accept “nitrogen AND electron lost closer to the nucleus «than magnesium»”.

e(iv).

Any two of:

subatomic particles «discovered»
OR
particles smaller/with masses less than atoms «discovered»
OR
«existence of» isotopes «same number of protons, different number of neutrons» ✔

charged particles obtained from «neutral» atoms
OR
atoms can gain or lose electrons «and become charged» ✔

atom «discovered» to have structure ✔

fission
OR
atoms can be split ✔

Accept atoms can undergo fusion «to produce heavier atoms»

Accept specific examples of particles.

Award [2] for “atom shown to have a nucleus with electrons around it” as both M1 and M3.

Award [1] for all bonding types correct.

Award [1] for each correct description.

Apply ECF for M2 only once.

Examiners report

Done very well. However, it was disappointing to see the formula of oxygen molecule as O and the oxide as Mg₂O and MgO₂ at HL level.

a(i).

Average performance; the question asked to identify a metal; however, answers included S, Si, P and even noble gases besides Be and Na. The only choice of aluminium; however, since its oxide is amphoteric, it could not be the answer in the minds of some.

a(ii).

Very good performance; some calculated the mass of oxygen instead of magnesium for the calculation of the amount, in mol, of magnesium. Others calculated the mass, but not the amount in mol as required.

b(i).

Mediocre performance; instead of calculating percentage uncertainty, some calculated percentage difference.

b(ii).

Satisfactory performance; however, a good number could not answer the question correctly on determining the percentage yield.

b(iii).

Poorly done. The question asked to evaluate and explain but instead many answers simply agreed with the information provided instead of assessing its strength and limitation.

c(i).

Mediocre performance; explaining the yield found was often a challenge by not recognizing that incomplete reaction or Mg partially oxidized or impurities present that evaporated or did not react would explain the yield.

c(ii).

Calculating coefficients that balance the given equation was done very well.

d(i).

Well done; some chose bromocresol green or methyl red as the indicator that would change colour, instead of phenol red, bromothymol blue or phenolphthalein.

d(ii).

Good performance; however, surprising number of candidates could not determine one or both oxidation states correctly or wrote it as 3 or 3−, instead of −3.

d(iii).

Average performance; choosing the given reaction as an acid-base or redox reaction was not done well. Often answers were contradictory and the reasoning incorrect.

d(iv).

Stating the number of subatomic particles in a ¹⁴N^3- was done very well. However, some answers showed a lack of understanding of how to calculate the number of relevant subatomic particles given formula of an ion with charge and mass number.

e(i).

Exceptionally well done; A few candidates referred to isomers, rather than isotopes.

e(ii).

There was reference to nitrogen and magnesium, rather than nitride and magnesium ions. Also, instead identifying smaller nuclear charge in nitride ion, some referred to core electrons, Z_eff, increased electron-electron repulsion or shielding.

e(iii).

Common error in suggesting nitrogen would have the greater sixth ionization energy was that for nitrogen, electron is lost from first energy level without making reference to magnesium losing it from second energy level.

e(iv).

Good performance; some teachers were concerned about the expected answers. However, generally, students were able to suggest two reasons why matter is divisible.

One teacher commented that not asking to describe bonding in terms of electrostatic attractions as in earlier papers would have been confusing and some did answer in terms of electrostatic forces of attractions involved. However, the question was clear in its expectation that the answer had to be in terms of how the valence electrons produce the three types of bonds and the overall performance was good. Some had difficulty identifying the bond type for Mg, O₂ and MgO.

Syllabus sections

Core » Topic 1: Stoichiometric relationships » 1.3 Reacting masses and volumes

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